Answer to Question #168461 in Mechanics | Relativity for April

Question

 A 3.5-kg block rest on a horizontal tabletop. A cord is attached to this block, the cord passing 

over a pulley at the edge of the table. At the end of the cord is hung another block weighing 

5 N. Determine the tension in the cord and the distance the blocks moved after 2 seconds.

Answer 1

The distance the blocks moved is:

$d=v_0t+\frac{at^2}{2}=\frac{at^2}{2},$

where $v_0$ - the initial speed in this case equal to zero;

a - the acceleration of the blocks.

Suppose the tabletop is frictionless, then:

$m_1a=T,$

$m_2a=m_2g-T,$

$m_2a=m_2g-m_1a,$

$a=\frac{m_2g}{m_1+m_2}=\frac{G_2}{m_1+\frac{G_2}{g}},$

where T - the tension in the cord;

$m_1$ - the mass of the first block;

$G_2$ - the weight of the second block.

$d=\frac{G_2t^2}{2(m_1+\frac{G_2}{g})}=\frac{5\cdot 2^2}{2(3.5+\frac{5}{9.8})}\approx 2.5\space m.$

The tension in the cord is:

$T=m_1a=\frac{m_1G_2}{m_1+\frac{G_2}{g}}=\frac{3.5\cdot5 }{3.5+\frac{5}{9.8}}\approx 4.4\space N.$

Answer: 4.4 N; 2.5 m.